$\int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) d x=$

The integral $I = \int_{\log \frac{1}{2}}^{\log 2} \sin \left(\frac{e^x-1}{e^x+1}\right) dx$ has limits $\log \frac{1}{2} = -\log 2$ and $\log 2$, forming the symmetric interval $[-\log 2, \log 2]$.

Consider the function $f(x) = \sin \left(\frac{e^x-1}{e^x+1}\right)$. Substituting $-x$ gives $f(-x) = \sin \left(\frac{e^{-x}-1}{e^{-x}+1}\right)$. Multiplying numerator and denominator by $e^x$ yields $f(-x) = \sin \left(\frac{1-e^x}{1+e^x}\right) = \sin \left(-\frac{e^x-1}{e^x+1}\right) = -\sin \left(\frac{e^x-1}{e^x+1}\right) = -f(x)$.

Since $f(x)$ is an odd function and the interval is symmetric about zero, the definite integral vanishes: $I = 0$.

Final Answer: $\boxed{0}$